Post by comp.soft-sys.matlab....I try to proove an equation but I faced some difficulties for proovingfor n odd valuessum i=1 to ((n-1)/2) [cos((2*i-1)*pi/n)] = 0,5in other words,cos(pi/n)+cos(3*pi/n)+cos(5*pi/n)+.............+cos((n-2)*pi/n) = 0,5....

You might have been answered more promptly in a news group such as<sci.math> which carries much more traffic than <aus.mathematics>. Butsince nobody else has replied yet, here are a couple of methods to proveyour equation. (Notice "prove", not "proove". :-)

The obvious trick is to use complex numbers. If we write

C = (sum i=1 to (n-1)/2) (cos((2i-1)pi/n)) and also

S = (sum i=1 to (n-1)/2) (sin((2i-1)pi/n))

then C + iS is a geometric series by de Moivre's Theorem, so it caneasily be summed. However, the sum then needs some slightly trickycancellation and simplification before you can find its real part C.

A shorter method uses the less well-known trick of multiplyingboth sides by 2.sin(pi/n) and using the prosthaphaeresis formula

Post by comp.soft-sys.matlab....I try to proove an equation but I faced some difficulties for proovingfor n odd valuessum i=1 to ((n-1)/2) [cos((2*i-1)*pi/n)] = 0,5in other words,cos(pi/n)+cos(3*pi/n)+cos(5*pi/n)+.............+cos((n-2)*pi/n) = 0,5....

You might have been answered more promptly in a news group such as<sci.math> which carries much more traffic than <aus.mathematics>. Butsince nobody else has replied yet, here are a couple of methods to proveyour equation. (Notice "prove", not "proove". :-)

I was assuming that it was homework -- certainly it's a standardenough kind of question. Still, as long as you've (Ken) put ananswer out there, here's another approach, which may hopefully beinstructive -- essentially, we restore some missing symmetry to theequation.

Draw a unit circle, and mark the points corresponding to each ofthe angles in the above sum. These form part of a regular n-gon;the remaining points in this n-gon are easily seen to arise from piand the negations of the original angles. Now consider the sum ofthe cosines of the angles in the entire n-gon. If your originalsum is S1, then since Cos(-x) = Cos(x) this new sum (call it S2)satisfies S2 = 2*S1 - 1.

Treating the points as complex numbers, it is clear that they arethe roots of x^n + 1 = 0. By elementary symmetric function theory,the sum of the roots is the negative of the coefficient of x^(n-1),which is 0. S2 is the real part of this sum, thus S2 = 0, henceS1 = 1/2.

Alternatively, one could use a vector argument (really, it's thesame approach, but may feel more natural if one is not comfortablewith the above) -- S2 is the x value of the sum of the vectors inthe n-gon. But a straightforward rearrangement turns these vectorsinto the sides of a regular n-gon, thus their sum is zero and theresult follows as before.

No doubt there are other ways to proceed, but what I feel is thekey point here is that rewriting the equation in a more symmetricform (as a function over the vertices of a regular n-gon) greatlyhelps simplify matters. Exploiting symmetry is a very powerfulmathematical tool.

Post by comp.soft-sys.matlabHello everybody,I try to proove an equation but I faced some difficulties for proovingfor n odd valuessum i=1 to ((n-1)/2) [cos((2*i-1)*pi/n)] = 0,5in other words,cos(pi/n)+cos(3*pi/n)+cos(5*pi/n)+.............+cos((n-2)*pi/n) = 0,5Thanks in advance, Best Regards

============Let n=2m+1 and

(1) T_m(x)=cos(m*arccos(x)) =(Chebychev-polynomial)and

(2) Q_m(x)=(T_m(x)+T_{m+1}(x))/(1+x) .

Because T_m(-1)=(-1)^m observe that Q_m(x) is a polynomial of degreem.